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Given a $n \times n$ matrix, the $(i, j)$ minor is the determinant of the submatrix formed by deleting the i-th row and j-th column. If the sum of all row vectors and the sum of all column vectors are both 0, then the $(i, j)$ minor does not depend on the choice of $i$ and $j$ (up to $\pm 1$). In knot theory, this is used to define the determinate of a knot via coloring matrix.

My question is, can we find a necessary and sufficient condition for the matrix such that the absolute values of all $(n-1)\times (n-1)$ minors of it are equal? Or some other sufficient conditions besides the above one?

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  • $\begingroup$ If the we assume that each $(i, j)$ minor $\neq 0$, then all cofactors are equivelent if and only if the sum of all rows and the sum of all columns are both 0. That is the result I want. It seems much easier than I imagined : ) $\endgroup$ Commented Oct 25, 2013 at 7:00

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